Question

Express $-1+i/\sqrt{2}$ in polar form.

Answer 1

$z=-1+\frac{1}{\sqrt{2}}$

$r=\sqrt{1+{\left(\frac{1}{\sqrt{2}}\right)}^2}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$

$a=r\cos \theta b=r\sin \theta$

$-1=\sqrt{\frac{3}{2}}\cos \theta \frac{1}{\sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}\sin \theta$

$\theta ={\cos }^{-1}\left(\sqrt{\frac{2}{3}}\right)\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\therefore z=r(\cos \theta +\sin \theta )=\sqrt{\frac{3}{2}}(\sqrt{\frac{2}{3}}+1\left(\frac{1}{\sqrt{3}}\right))$

$z=\sqrt{\frac{3}{2}}(\sqrt{\frac{2}{3}}+1\left(\frac{1}{\sqrt{3}}\right))$